Calculate output voltage and current of a boost converter

If I have two AA rechargeable battery (1.2 V, 2000 mAh) connected in series, and I then connect it to a boost converter (DC to DC converter) so that the output voltage is 5V, what is the maximum current I can get? Can you please show how to calculate this?

Just to give you a bit more info, I have one of these so called "mobile boosters" which can charge your phone from 2 AA batteries. At the back, there is a small label saying that the maximum output is 5 V @ 500 mA . However when I plug it to my Blackberry, I turn on the engineering screen, I can see that the charging current is 1250 mA which is what I normally get when I use the wall charger. When I plug it to a computer, the charging current become 500 mA which makes sense as USB2 port can supply up to 500 mA. What I don't get is why does it say 1250 mA when the source can only supply 500 mA. So is the Blackberry gives out false information or the mobile booster actually deliver more than 500 mA?

Please try to keep the answer simple, don't get too technical, I'm not that great with physics

If the mobile booster doesn't have a regulator, it will output an amount equal to the load that is on it, until it reaches a point where it can no longer match the load (over loaded) or it's source power can no longer match the load (the batteries start going dead). As you are seeing a higher output than what the sticker on it says, I suspect it either doesn't have a regulator or if it does it isn't what the sticker is claiming.
The BlackBerry device has a regulator built in, so it will show whatever it's load capacity is minus the load capacity of the charge source. If both are the same, then it will show whatever it's max is.
If the source is less, it will show less than it's max. If the source is more, the regulator will step it down and it will just show the max.

In simplified numbers, for a switched-mode boost converter, assuming N=efficiency in percent and I=current:

Vin x Iin = N (Vout x Iout)

N = 0.95 for an efficient boost converter

Rearranging for Iout, into the 3.8 V typical battery voltage (rather than the rated 5 V):

Iout = Vin / Vout * Iin / N

Iout = (1.2 V * 2) / 3.8 V * 2000 mA / 0.95

Iout = 1.2 A

This is similar to the charge current value you are seeing, but several things should be noted, along the lines of what DaFoxGrey pointed out:
- the output voltage is likely unregulated, but the output capacity is specified at 5 V as a standard, even though it will never rise above 3.8 V to 4.2 V when charging a Lithium Ion battery.
- the output charge current will be directly related to the difference between the theoretical converter output voltage and the present battery voltage, so Iout will be highest when Vbat is low and vice-versa as Vbat rises towards 5 V as it charges.
- the output charge current and instantaneous voltage is highly dependent on the condition of the input batteries, as well as the current state of the phone battery.
- Li-Ion charge circuits typically contain protection (which will de-rate or shut off output current) against overcharging , which can permanently damage or overheat the battery.